Bezier vs B-Spline
(Supplement Notes)
Monday, February 18, 13
Bezier curve
• Developed by Paul de Casteljau (1959) and independently by Pierre Bezier (1962).
• French automobil company – Citroen & Renault.
P0
P1 P2
P3
Monday, February 18, 13
Parametric function
• P(u) = ∑ Bn,i(u)pi Where Bn,i(u) = . n!. ui(1-u)n-i
i!(n-i)! 0<= u<= 1
i=0
n
Monday, February 18, 13
Parametric function
• P(u) = ∑ Bn,i(u)pi Where Bn,i(u) = . n!. ui(1-u)n-i
i!(n-i)! 0<= u<= 1
i=0
n
For 3 control points, n = 2
Monday, February 18, 13
Parametric function
• P(u) = ∑ Bn,i(u)pi Where Bn,i(u) = . n!. ui(1-u)n-i
i!(n-i)! 0<= u<= 1
i=0
n
For 3 control points, n = 2P(u) = (1-u)2p0 + 2u(1-u) p1+ u2p2
Monday, February 18, 13
Parametric function
• P(u) = ∑ Bn,i(u)pi Where Bn,i(u) = . n!. ui(1-u)n-i
i!(n-i)! 0<= u<= 1
i=0
n
For 3 control points, n = 2P(u) = (1-u)2p0 + 2u(1-u) p1+ u2p2
For four control points, n = 3
Monday, February 18, 13
Parametric function
• P(u) = ∑ Bn,i(u)pi Where Bn,i(u) = . n!. ui(1-u)n-i
i!(n-i)! 0<= u<= 1
i=0
n
For 3 control points, n = 2P(u) = (1-u)2p0 + 2u(1-u) p1+ u2p2
For four control points, n = 3P(u) = (1-u)3p0 + 3u(1-u) 2 p1 + 3u 2 (1-u)p2 + u3p3
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
u=0.75
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
u=0.7520
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
u=0.7520
21
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
u=0.7520
21
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
u=0.7520
21
Monday, February 18, 13
algorithm
• De Casteljau– Basic concept
• To choose a point C in line segment AB such that C divides the line segment AB in a ratio of u: 1-u
A BC
P0
P1
P2
Let u = 0.5
00 01u=0.25
10
11
u=0.7520
21
Monday, February 18, 13
properties
Monday, February 18, 13
properties
• The curve passes through the first, P0 and last vertex points, Pn .
Monday, February 18, 13
properties
• The curve passes through the first, P0 and last vertex points, Pn .
• The tangent vector at the starting point P0 must be given by P1 – P0 and the tangent Pn given by Pn – Pn-1
Monday, February 18, 13
properties
• The curve passes through the first, P0 and last vertex points, Pn .
• The tangent vector at the starting point P0 must be given by P1 – P0 and the tangent Pn given by Pn – Pn-1
• This requirement is generalized for higher derivatives at the curve’s end points. E.g 2nd derivative at P0 can be determined by P0 ,P1 ,P2 (to satisfy continuity)
Monday, February 18, 13
properties
• The curve passes through the first, P0 and last vertex points, Pn .
• The tangent vector at the starting point P0 must be given by P1 – P0 and the tangent Pn given by Pn – Pn-1
• This requirement is generalized for higher derivatives at the curve’s end points. E.g 2nd derivative at P0 can be determined by P0 ,P1 ,P2 (to satisfy continuity)
• The same curve is generated when the order of the control points is reversed
Monday, February 18, 13
Properties (continued)
• Convex hull– Convex polygon formed by connecting the
control points of the curve.– Curve resides completely inside its convex hull
Monday, February 18, 13
B-Spline• Motivation (recall bezier curve)
– The degree of a Bezier Curve is determined by the number of control points
– E. g. (bezier curve degree 11) – difficult to bend the "neck" toward the line segment P4P5.
– Of course, we can add more control points.
– BUT this will increase the degree of the curve à increase computational burden
Monday, February 18, 13
• Motivation (recall bezier curve)– Joint many bezier curves of lower
degree together (right figure)– BUT maintaining continuity in
the derivatives of the desired order at the connection point is not easy or may be tedious and undesirable.
B-Spline
Monday, February 18, 13
B-Spline
• Motivation (recall bezier curve)– moving a control point affects the
shape of the entire curve- (global modification property) – undesirable.
- Thus, the solution is B-Spline – the degree of the curve is independent of the number of control points
- E.g - right figure – a B-spline curve of degree 3 defined by 8 control points
Monday, February 18, 13
• In fact, there are five Bézier curve segments of degree 3 joining together to form the B-spline curve defined by the control points
• little dots subdivide the B-spline curve into Bézier curve segments.
• Subdividing the curve directly is difficult to do à so, subdivide the domain of the curve by points called knots
B-Spline
0 u 1
Monday, February 18, 13
• In summary, to design a B-spline curve, we need a set of control points, a set of knots and a degree of curve.
B-Spline
Monday, February 18, 13
• P(u) = ∑ Ni,k(u)pi (u0 < u < um).. (1.0)
Where basis function = Ni,k(u)Degree of curve à k-1Control points, pi à 0 < i < nKnot, u à u0 < u < um
m = n + k
B-Spline curve
i=0
n
Monday, February 18, 13
B-Spline : definition
• P(u) = ∑ Ni,k(u)pi (u0 < u < um)• ui à knot• [ui, ui+1) à knot span• (u0, u1, u2, …. um )à knot vector• The point on the curve that corresponds to a knot ui, à
knot point ,P(ui)• If knots are equally space à uniform (e.g, 0, 0.2, 0.4,
0.6…)• Otherwise à non uniform (e.g: 0, 0.1, 0.3, 0.4, 0.8 …)
Monday, February 18, 13
• Uniform knot vector– Individual knot value is evenly spaced – (0, 1, 2, 3, 4)– Then, normalized to the range [0, 1]– (0, 0.25, 0.5, 0.75, 1.0)
B-Spline : definition
Monday, February 18, 13
Type of B-Spline uniform knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
Monday, February 18, 13
Type of B-Spline uniform knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
-First and last knots are duplicated k times.
Monday, February 18, 13
Type of B-Spline uniform knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
-First and last knots are duplicated k times.-E.g (0,0,0,1,2,2,2)
Monday, February 18, 13
Type of B-Spline uniform knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
-First and last knots are duplicated k times.-E.g (0,0,0,1,2,2,2)-Curve pass through the first and last control points
Monday, February 18, 13
Type of B-Spline uniform knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
-First and last knots are duplicated k times.-E.g (0,0,0,1,2,2,2)-Curve pass through the first and last control points
-First and last knots are not duplicated – same contribution.
Monday, February 18, 13
Type of B-Spline uniform knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
-First and last knots are duplicated k times.-E.g (0,0,0,1,2,2,2)-Curve pass through the first and last control points
-First and last knots are not duplicated – same contribution.-E.g (0, 1, 2, 3)
Monday, February 18, 13
Type of B-Spline uniform knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
-First and last knots are duplicated k times.-E.g (0,0,0,1,2,2,2)-Curve pass through the first and last control points
-First and last knots are not duplicated – same contribution.-E.g (0, 1, 2, 3)-Curve doesn’t pass through end points.
Monday, February 18, 13
Type of B-Spline uniform knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
-First and last knots are duplicated k times.-E.g (0,0,0,1,2,2,2)-Curve pass through the first and last control points
-First and last knots are not duplicated – same contribution.-E.g (0, 1, 2, 3)-Curve doesn’t pass through end points.- used to generate closed curves (when first = last control points)
Monday, February 18, 13
Type of B-Spline knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
Monday, February 18, 13
Type of B-Spline knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
Monday, February 18, 13
Type of B-Spline knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
Monday, February 18, 13
Type of B-Spline knot vector
Non-periodic knots (open knots)
Periodic knots (non-open knots)
(Closed knots)
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Non-periodic (open) uniform B-Spline• The knot spacing is evenly spaced except at the ends
where knot values are repeated k times.
i=0
n
Monday, February 18, 13
Non-periodic (open) uniform B-Spline• The knot spacing is evenly spaced except at the ends
where knot values are repeated k times.• E.g P(u) = ∑ Ni,k(u)pi (u0 < u < um)
i=0
n
Monday, February 18, 13
Non-periodic (open) uniform B-Spline• The knot spacing is evenly spaced except at the ends
where knot values are repeated k times.• E.g P(u) = ∑ Ni,k(u)pi (u0 < u < um)• Degree = k-1, number of control points = n + 1
i=0
n
Monday, February 18, 13
Non-periodic (open) uniform B-Spline• The knot spacing is evenly spaced except at the ends
where knot values are repeated k times.• E.g P(u) = ∑ Ni,k(u)pi (u0 < u < um)• Degree = k-1, number of control points = n + 1• Number of knots = m + 1 @ n+ k + 1
i=0
n
Monday, February 18, 13
Non-periodic (open) uniform B-Spline• The knot spacing is evenly spaced except at the ends
where knot values are repeated k times.• E.g P(u) = ∑ Ni,k(u)pi (u0 < u < um)• Degree = k-1, number of control points = n + 1• Number of knots = m + 1 @ n+ k + 1àfor degree = 1 and number of control points = 4 à(k = 2, n = 3)
i=0
n
Monday, February 18, 13
Non-periodic (open) uniform B-Spline• The knot spacing is evenly spaced except at the ends
where knot values are repeated k times.• E.g P(u) = ∑ Ni,k(u)pi (u0 < u < um)• Degree = k-1, number of control points = n + 1• Number of knots = m + 1 @ n+ k + 1àfor degree = 1 and number of control points = 4 à(k = 2, n = 3)à Number of knots = n + k + 1 = 6
i=0
n
Monday, February 18, 13
Non-periodic (open) uniform B-Spline• The knot spacing is evenly spaced except at the ends
where knot values are repeated k times.• E.g P(u) = ∑ Ni,k(u)pi (u0 < u < um)• Degree = k-1, number of control points = n + 1• Number of knots = m + 1 @ n+ k + 1àfor degree = 1 and number of control points = 4 à(k = 2, n = 3)à Number of knots = n + k + 1 = 6 non periodic uniform knot vector (0,0,1,2,3, 3)
i=0
n
Monday, February 18, 13
Non-periodic (open) uniform B-Spline• The knot spacing is evenly spaced except at the ends
where knot values are repeated k times.• E.g P(u) = ∑ Ni,k(u)pi (u0 < u < um)• Degree = k-1, number of control points = n + 1• Number of knots = m + 1 @ n+ k + 1àfor degree = 1 and number of control points = 4 à(k = 2, n = 3)à Number of knots = n + k + 1 = 6 non periodic uniform knot vector (0,0,1,2,3, 3)* Knot value between 0 and 3 are equally spaced à
uniform
i=0
n
Monday, February 18, 13
• Example• For curve degree = 3, number of control points = 5• à k = 4, n = 4• à number of knots = n+k+1 = 9• à non periodic knots vector = (0,0,0,0,1,2,2,2)• For curve degree = 1, number of control points = 5• à k = 2, n = 4• à number of knots = n + k + 1 = 7• à non periodic uniform knots vector = (0, 0, 1, 2, 3, 4, 4)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
• For any value of parameters k and n, non periodic knots are determined from
Non-periodic (open) uniform B-Spline
(1.3)
Monday, February 18, 13
• For any value of parameters k and n, non periodic knots are determined from
Non-periodic (open) uniform B-Spline
ui = 0 0≤ i < ki – k + 1 k ≤ i ≤ nn – k + 2 n < i ≤n+k
(1.3)
Monday, February 18, 13
• For any value of parameters k and n, non periodic knots are determined from
Non-periodic (open) uniform B-Spline
ui = 0 0≤ i < ki – k + 1 k ≤ i ≤ nn – k + 2 n < i ≤n+k
e.g k=2, n = 3
(1.3)
Monday, February 18, 13
• For any value of parameters k and n, non periodic knots are determined from
Non-periodic (open) uniform B-Spline
ui = 0 0≤ i < ki – k + 1 k ≤ i ≤ nn – k + 2 n < i ≤n+k
e.g k=2, n = 3
ui = 0 0≤ i < 2i – 2 + 1 2 ≤ i ≤ 33 – 2 + 2 3 < i ≤5
(1.3)
Monday, February 18, 13
• For any value of parameters k and n, non periodic knots are determined from
Non-periodic (open) uniform B-Spline
ui = 0 0≤ i < ki – k + 1 k ≤ i ≤ nn – k + 2 n < i ≤n+k
e.g k=2, n = 3
ui = 0 0≤ i < 2i – 2 + 1 2 ≤ i ≤ 33 – 2 + 2 3 < i ≤5
u = (0, 0, 1, 2, 3, 3)
(1.3)
Monday, February 18, 13
B-Spline basis function
àIn equation (1.1), the denominators can have a value of zero, 0/0 is presumed to be zero.
àIf the degree is zero basis function Ni,1(u) is 1 if u is in the i-th knot span [ui, ui+1).
Otherwise
(1.1)
(1.2)
Monday, February 18, 13
• For example, if we have four knots u0 = 0, u1 = 1, u2 = 2 and u3 = 3, knot spans 0, 1 and 2 are [0,1), [1,2), [2,3)
• the basis functions of degree 0 are N0,1(u) = 1 on [0,1) and 0 elsewhere, N1,1(u) = 1 on [1,2) and 0 elsewhere, and N2,1(u) = 1 on [2,3) and 0 elsewhere.
• This is shown below
B-Spline basis function
Monday, February 18, 13
• To understand the way of computing Ni,p(u) for p greater than 0, we use the triangular computation scheme
B-Spline basis function
Monday, February 18, 13
Example• Find the knot values of a non periodic
uniform B-Spline which has degree = 2 and 3 control points. Then, find the equation of B-Spline curve in polynomial form.
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer• Degree = k-1 = 2 à k=3
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer• Degree = k-1 = 2 à k=3• Control points = n + 1 = 3 à n=2
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer• Degree = k-1 = 2 à k=3• Control points = n + 1 = 3 à n=2• Number of knot = n + k + 1 = 6
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer• Degree = k-1 = 2 à k=3• Control points = n + 1 = 3 à n=2• Number of knot = n + k + 1 = 6• Knot values à u0=0, u1=0, u2=0, u3=1,u4=1,u5= 1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
i=0
n
i=0
2
Monday, February 18, 13
Answer(cont)
Non-periodic (open) uniform B-Spline
i=0
n
i=0
2
Monday, February 18, 13
Answer(cont)• To obtain the polynomial equation,
P(u) = ∑ Ni,k(u)pi
Non-periodic (open) uniform B-Spline
i=0
n
i=0
2
Monday, February 18, 13
Answer(cont)• To obtain the polynomial equation,
P(u) = ∑ Ni,k(u)pi
• = ∑ Ni,3(u)pi
Non-periodic (open) uniform B-Spline
i=0
n
i=0
2
Monday, February 18, 13
Answer(cont)• To obtain the polynomial equation,
P(u) = ∑ Ni,k(u)pi
• = ∑ Ni,3(u)pi
• = N0,3(u)p0 + N1,3(u)p1 + N2,3(u)p2
Non-periodic (open) uniform B-Spline
i=0
n
i=0
2
Monday, February 18, 13
Answer(cont)• To obtain the polynomial equation,
P(u) = ∑ Ni,k(u)pi
• = ∑ Ni,3(u)pi
• = N0,3(u)p0 + N1,3(u)p1 + N2,3(u)p2
• firstly, find the Ni,k(u) using the knot value that shown above, start from k =1 to k=3
Non-periodic (open) uniform B-Spline
i=0
n
i=0
2
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise• N1,1(u) = 1 u1≤ u ≤u2 ; (u=0)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise• N1,1(u) = 1 u1≤ u ≤u2 ; (u=0)• 0 otherwise
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise• N1,1(u) = 1 u1≤ u ≤u2 ; (u=0)• 0 otherwise• N2,1(u) = 1 u2≤ u ≤u3 ; (0≤ u ≤ 1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise• N1,1(u) = 1 u1≤ u ≤u2 ; (u=0)• 0 otherwise• N2,1(u) = 1 u2≤ u ≤u3 ; (0≤ u ≤ 1)• 0 otherwise
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise• N1,1(u) = 1 u1≤ u ≤u2 ; (u=0)• 0 otherwise• N2,1(u) = 1 u2≤ u ≤u3 ; (0≤ u ≤ 1)• 0 otherwise• N3,1(u) = 1 u3≤ u ≤u4 ; (u=1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise• N1,1(u) = 1 u1≤ u ≤u2 ; (u=0)• 0 otherwise• N2,1(u) = 1 u2≤ u ≤u3 ; (0≤ u ≤ 1)• 0 otherwise• N3,1(u) = 1 u3≤ u ≤u4 ; (u=1)• 0 otherwise
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise• N1,1(u) = 1 u1≤ u ≤u2 ; (u=0)• 0 otherwise• N2,1(u) = 1 u2≤ u ≤u3 ; (0≤ u ≤ 1)• 0 otherwise• N3,1(u) = 1 u3≤ u ≤u4 ; (u=1)• 0 otherwise• N4,1(u) = 1 u4≤ u ≤u5 ; (u=1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)• 0 otherwise• N1,1(u) = 1 u1≤ u ≤u2 ; (u=0)• 0 otherwise• N2,1(u) = 1 u2≤ u ≤u3 ; (0≤ u ≤ 1)• 0 otherwise• N3,1(u) = 1 u3≤ u ≤u4 ; (u=1)• 0 otherwise• N4,1(u) = 1 u4≤ u ≤u5 ; (u=1)• 0 otherwise
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =u2 = 0)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =u2 = 0) • u1 - u0 u2 – u1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =u2 = 0) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 0 – u N1,1 = 0
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =u2 = 0) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 0 – u N1,1 = 0• 0 – 0 0 – 0
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =u2 = 0) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 0 – u N1,1 = 0• 0 – 0 0 – 0 • N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =u2 = 0, u3 = 1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =u2 = 0) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 0 – u N1,1 = 0• 0 – 0 0 – 0 • N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =u2 = 0, u3 = 1)
• u2 - u1 u3 – u2
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =u2 = 0) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 0 – u N1,1 = 0• 0 – 0 0 – 0 • N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =u2 = 0, u3 = 1)
• u2 - u1 u3 – u2
• = u – 0 N1,1 + 1 – u N2,1 = 1 - u
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =u2 = 0) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 0 – u N1,1 = 0• 0 – 0 0 – 0 • N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =u2 = 0, u3 = 1)
• u2 - u1 u3 – u2
• = u – 0 N1,1 + 1 – u N2,1 = 1 - u• 0 – 0 1 – 0
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =0, u3 =u4 = 1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =0, u3 =u4 = 1) • u3 – u2 u4 – u3
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =0, u3 =u4 = 1) • u3 – u2 u4 – u3
• = u – 0 N2,1 + 1 – u N3,1 = u
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =0, u3 =u4 = 1) • u3 – u2 u4 – u3
• = u – 0 N2,1 + 1 – u N3,1 = u• 1 – 0 1 – 1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =0, u3 =u4 = 1) • u3 – u2 u4 – u3
• = u – 0 N2,1 + 1 – u N3,1 = u• 1 – 0 1 – 1 • N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 =u4 = u5 = 1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =0, u3 =u4 = 1) • u3 – u2 u4 – u3
• = u – 0 N2,1 + 1 – u N3,1 = u• 1 – 0 1 – 1 • N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 =u4 = u5 = 1)
• u4 – u3 u5– u4
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =0, u3 =u4 = 1) • u3 – u2 u4 – u3
• = u – 0 N2,1 + 1 – u N3,1 = u• 1 – 0 1 – 1 • N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 =u4 = u5 = 1)
• u4 – u3 u5– u4
• = u – 1 N3,1 + 1 – u N4,1 = 0
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =0, u3 =u4 = 1) • u3 – u2 u4 – u3
• = u – 0 N2,1 + 1 – u N3,1 = u• 1 – 0 1 – 1 • N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 =u4 = u5 = 1)
• u4 – u3 u5– u4
• = u – 1 N3,1 + 1 – u N4,1 = 0• 1 – 1 1 – 1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-SplineAnswer (cont)For k = 2N0,2(u) = 0N1,2(u) = 1 - uN2,2(u) = uN3,2(u) = 0
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
• N0,3(u) = u - u0 N0,2 + u3 – u N1,2 (u0 =u1 =u2 = 0, u3 =1 )
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
• N0,3(u) = u - u0 N0,2 + u3 – u N1,2 (u0 =u1 =u2 = 0, u3 =1 )
• u2 - u0 u3 – u1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
• N0,3(u) = u - u0 N0,2 + u3 – u N1,2 (u0 =u1 =u2 = 0, u3 =1 )
• u2 - u0 u3 – u1
• = u – 0 N0,2 + 1 – u N1,2 = (1-u)(1-u) = (1- u)2
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
• N0,3(u) = u - u0 N0,2 + u3 – u N1,2 (u0 =u1 =u2 = 0, u3 =1 )
• u2 - u0 u3 – u1
• = u – 0 N0,2 + 1 – u N1,2 = (1-u)(1-u) = (1- u)2
• 0 – 0 1 – 0
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
• N0,3(u) = u - u0 N0,2 + u3 – u N1,2 (u0 =u1 =u2 = 0, u3 =1 )
• u2 - u0 u3 – u1
• = u – 0 N0,2 + 1 – u N1,2 = (1-u)(1-u) = (1- u)2
• 0 – 0 1 – 0 • N1,3(u) = u - u1 N1,2 + u4 – u N2,2 (u1 =u2 = 0, u3 = u4 =
1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
• N0,3(u) = u - u0 N0,2 + u3 – u N1,2 (u0 =u1 =u2 = 0, u3 =1 )
• u2 - u0 u3 – u1
• = u – 0 N0,2 + 1 – u N1,2 = (1-u)(1-u) = (1- u)2
• 0 – 0 1 – 0 • N1,3(u) = u - u1 N1,2 + u4 – u N2,2 (u1 =u2 = 0, u3 = u4 =
1)• u3 - u1 u4 – u2
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
• N0,3(u) = u - u0 N0,2 + u3 – u N1,2 (u0 =u1 =u2 = 0, u3 =1 )
• u2 - u0 u3 – u1
• = u – 0 N0,2 + 1 – u N1,2 = (1-u)(1-u) = (1- u)2
• 0 – 0 1 – 0 • N1,3(u) = u - u1 N1,2 + u4 – u N2,2 (u1 =u2 = 0, u3 = u4 =
1)• u3 - u1 u4 – u2
• = u – 0 N1,2 + 1 – u N2,2 = u(1 – u) +(1-u)u = 2u(1-u)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 3, find Ni,3(u) – use equation (1.1):
• N0,3(u) = u - u0 N0,2 + u3 – u N1,2 (u0 =u1 =u2 = 0, u3 =1 )
• u2 - u0 u3 – u1
• = u – 0 N0,2 + 1 – u N1,2 = (1-u)(1-u) = (1- u)2
• 0 – 0 1 – 0 • N1,3(u) = u - u1 N1,2 + u4 – u N2,2 (u1 =u2 = 0, u3 = u4 =
1)• u3 - u1 u4 – u2
• = u – 0 N1,2 + 1 – u N2,2 = u(1 – u) +(1-u)u = 2u(1-u)• 1 – 0 1 – 0
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)• N2,3(u) = u – u2 N2,2 + u5 – u N3,2 (u2 =0, u3 =u4 = u5 =1)
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)• N2,3(u) = u – u2 N2,2 + u5 – u N3,2 (u2 =0, u3 =u4 = u5 =1) • u4 – u2 u5 – u3
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)• N2,3(u) = u – u2 N2,2 + u5 – u N3,2 (u2 =0, u3 =u4 = u5 =1) • u4 – u2 u5 – u3
• = u – 0 N2,2 + 1 – u N3,2 = u2
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)• N2,3(u) = u – u2 N2,2 + u5 – u N3,2 (u2 =0, u3 =u4 = u5 =1) • u4 – u2 u5 – u3
• = u – 0 N2,2 + 1 – u N3,2 = u2
• 1 – 0 1 – 1
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)• N2,3(u) = u – u2 N2,2 + u5 – u N3,2 (u2 =0, u3 =u4 = u5 =1) • u4 – u2 u5 – u3
• = u – 0 N2,2 + 1 – u N3,2 = u2
• 1 – 0 1 – 1 N0,3(u) =(1- u)2, N1,3(u) = 2u(1-u), N2,3(u) = u2
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)• N2,3(u) = u – u2 N2,2 + u5 – u N3,2 (u2 =0, u3 =u4 = u5 =1) • u4 – u2 u5 – u3
• = u – 0 N2,2 + 1 – u N3,2 = u2
• 1 – 0 1 – 1 N0,3(u) =(1- u)2, N1,3(u) = 2u(1-u), N2,3(u) = u2
The polynomial equation, P(u) = ∑ Ni,k(u)pi
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)• N2,3(u) = u – u2 N2,2 + u5 – u N3,2 (u2 =0, u3 =u4 = u5 =1) • u4 – u2 u5 – u3
• = u – 0 N2,2 + 1 – u N3,2 = u2
• 1 – 0 1 – 1 N0,3(u) =(1- u)2, N1,3(u) = 2u(1-u), N2,3(u) = u2
The polynomial equation, P(u) = ∑ Ni,k(u)pi• P(u) = N0,3(u)p0 + N1,3(u)p1 + N2,3(u)p2
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
Answer (cont)• N2,3(u) = u – u2 N2,2 + u5 – u N3,2 (u2 =0, u3 =u4 = u5 =1) • u4 – u2 u5 – u3
• = u – 0 N2,2 + 1 – u N3,2 = u2
• 1 – 0 1 – 1 N0,3(u) =(1- u)2, N1,3(u) = 2u(1-u), N2,3(u) = u2
The polynomial equation, P(u) = ∑ Ni,k(u)pi• P(u) = N0,3(u)p0 + N1,3(u)p1 + N2,3(u)p2
• = (1- u)2 p0 + 2u(1-u) p1 + u2p2 (0 <= u <= 1)
Non-periodic (open) uniform B-Spline
i=0
n
Monday, February 18, 13
• Exercise• Find the polynomial equation for curve with
degree = 1 and number of control points = 4
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
• Answer • k = 2 , n = 3 à number of knots = 6• Knot vector = (0, 0, 1, 2, 3, 3)• For k = 1, find Ni,1(u) – use equation (1.2):
• N0,1(u) = 1 u0≤ u ≤u1 ; (u=0)
• N1,1(u) = 1 u1≤ u ≤u2 ; (0≤ u ≤ 1) N2,1(u) = 1 u2≤ u ≤u3 ; (1≤ u ≤ 2)
• N3,1(u) = 1 u3≤ u ≤u4 ; (2≤ u ≤ 3) N4,1(u) = 1 u4≤ u ≤u5 ; (u=3)
•
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =0, u2 = 1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =0, u2 = 1) • u1 - u0 u2 – u1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =0, u2 = 1) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 1 – u N1,1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =0, u2 = 1) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 1 – u N1,1 • 0 – 0 1 – 0
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N0,2(u) = u - u0 N0,1 + u2 – u N1,1 (u0 =u1 =0, u2 = 1) • u1 - u0 u2 – u1
• = u – 0 N0,1 + 1 – u N1,1 • 0 – 0 1 – 0 • = 1 – u (0≤ u ≤ 1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =0, u2 =1, u3 = 2)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =0, u2 =1, u3 = 2)
• u2 - u1 u3 – u2
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =0, u2 =1, u3 = 2)
• u2 - u1 u3 – u2
• = u – 0 N1,1 + 2 – u N2,1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =0, u2 =1, u3 = 2)
• u2 - u1 u3 – u2
• = u – 0 N1,1 + 2 – u N2,1 • 1 – 0 2 – 1
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =0, u2 =1, u3 = 2)
• u2 - u1 u3 – u2
• = u – 0 N1,1 + 2 – u N2,1 • 1 – 0 2 – 1• N1,2(u) = u (0≤ u ≤ 1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• For k = 2, find Ni,2(u) – use equation (1.1):
• N1,2(u) = u - u1 N1,1 + u3 – u N2,1 (u1 =0, u2 =1, u3 = 2)
• u2 - u1 u3 – u2
• = u – 0 N1,1 + 2 – u N2,1 • 1 – 0 2 – 1• N1,2(u) = u (0≤ u ≤ 1)
• N1,2(u) = 2 – u (1≤ u ≤ 2)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =1, u3 =2,u4 = 3)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =1, u3 =2,u4 = 3) • u3 – u2 u4 – u3
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =1, u3 =2,u4 = 3) • u3 – u2 u4 – u3
• = u – 1 N2,1 + 3 – u N3,1 =
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =1, u3 =2,u4 = 3) • u3 – u2 u4 – u3
• = u – 1 N2,1 + 3 – u N3,1 =• 2 – 1 3 – 2
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =1, u3 =2,u4 = 3) • u3 – u2 u4 – u3
• = u – 1 N2,1 + 3 – u N3,1 =• 2 – 1 3 – 2 • N2,2(u) = u – 1 (1≤ u ≤ 2)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N2,2(u) = u – u2 N2,1 + u4 – u N3,1 (u2 =1, u3 =2,u4 = 3) • u3 – u2 u4 – u3
• = u – 1 N2,1 + 3 – u N3,1 =• 2 – 1 3 – 2 • N2,2(u) = u – 1 (1≤ u ≤ 2)
• N2,2(u) = 3 – u (2≤ u ≤ 3)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 = 2, u4 = 3, u5 =
3)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 = 2, u4 = 3, u5 =
3) • u4 – u3 u5– u4
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 = 2, u4 = 3, u5 =
3) • u4 – u3 u5– u4
• = u – 2 N3,1 + 3 – u N4,1 =
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 = 2, u4 = 3, u5 =
3) • u4 – u3 u5– u4
• = u – 2 N3,1 + 3 – u N4,1 = • 3 – 2 3 – 3
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• N3,2(u) = u – u3 N3,1 + u5 – u N4,1 (u3 = 2, u4 = 3, u5 =
3) • u4 – u3 u5– u4
• = u – 2 N3,1 + 3 – u N4,1 = • 3 – 2 3 – 3 • = u – 2 (2≤ u ≤ 3)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• The polynomial equation P(u) = ∑ Ni,k(u)pi
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• The polynomial equation P(u) = ∑ Ni,k(u)pi• P(u) = N0,2(u)p0 + N1,2(u)p1 + N2,2(u)p2 + N3,2(u)p3
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• The polynomial equation P(u) = ∑ Ni,k(u)pi• P(u) = N0,2(u)p0 + N1,2(u)p1 + N2,2(u)p2 + N3,2(u)p3
• P(u) = (1 – u) p0 + u p1 (0≤ u ≤ 1)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• The polynomial equation P(u) = ∑ Ni,k(u)pi• P(u) = N0,2(u)p0 + N1,2(u)p1 + N2,2(u)p2 + N3,2(u)p3
• P(u) = (1 – u) p0 + u p1 (0≤ u ≤ 1)
• P(u) = (2 – u) p1 + (u – 1) p2 (1≤ u ≤ 2)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Answer (cont)• The polynomial equation P(u) = ∑ Ni,k(u)pi• P(u) = N0,2(u)p0 + N1,2(u)p1 + N2,2(u)p2 + N3,2(u)p3
• P(u) = (1 – u) p0 + u p1 (0≤ u ≤ 1)
• P(u) = (2 – u) p1 + (u – 1) p2 (1≤ u ≤ 2)
• P(u) = (3 – u) p2 + (u - 2) p3 (2≤ u ≤ 3)
Non-periodic (open) uniform B-Spline
Monday, February 18, 13
Periodic uniform knot
• Periodic knots are determined from– Ui = i - k (0 ≤ i ≤ n+k)
• Example– For curve with degree = 3 and number of
control points = 4 (cubic B-spline)– (k = 4, n = 3) à number of knots = 8– (0, 1, 2, 3, 4, 5, 6, 8)
Monday, February 18, 13
• Normalize u (0<= u <= 1)• N0,4(u) = 1/6 (1-u)3
• N1,4(u) = 1/6 (3u 3 – 6u 2 +4)
• N2,4(u) = 1/6 (-3u 3 + 3u 2 + 3u +1)
• N3,4(u) = 1/6 u3
• P(u) = N0,4(u)p0 + N1,4(u)p1 + N2,4(u)p2 + N3,4(u)p3
Periodic uniform knot
Monday, February 18, 13
• In matrix form• P(u) = [u3,u2, u, 1].Mn.
• Mn = 1/6
Periodic uniform knot
P0P1P2P3
-1 3 -3 13 -6 3 0-3 0 3 01 4 1 0
Monday, February 18, 13
Periodic uniform knot
P0
Monday, February 18, 13
Closed periodic
P0
P1
P2
P3
P4
P5
Examplek = 4, n = 5
Monday, February 18, 13
Equation 1.0 change to • Ni,k(u) = N0,k((u-i)mod(n+1))
à P(u) = ∑ N0,k((u-i)mod(n+1))pi
Closed periodic
i=0
n
0<= u <= n+1
Monday, February 18, 13
Properties of B-Spline1. The m degree B-Spline function are
piecewise polynomials of degree m à have Cm-1 continuity. àe.g B-Spline degree 3 have C2 continuity.
u=1
u=2
Monday, February 18, 13
Properties of B-SplineIn general, the lower the degree, the closer a B-spline curve follows its control polyline.
Degree = 7 Degree = 5 Degree = 3
Monday, February 18, 13
Properties of B-Spline
Equality m = n + k must be satisfiedNumber of knots = m + 1k cannot exceed the number of control points, n+ 1
Monday, February 18, 13
Properties of B-Spline
2. Each curve segment is affected by k control points as shown by past examples.à e.g k = 3, P(u) = Ni-1,k pi-1 + Ni,k pi+ Ni+1,k pi+1
Monday, February 18, 13
Properties of B-SplineLocal Modification Scheme: changing the position of control point Pi only affects the curve C(u) on interval [ui, ui+k).
Modify control point P2
Monday, February 18, 13
Properties of B-Spline3. Strong Convex Hull Property: A B-spline curve is
contained in the convex hull of its control polyline. More specifically, if u is in knot span [ui,ui+1), then C(u) is in the convex hull of control points Pi-p, Pi-p+1, ..., Pi.
Degree = 3, k = 4Convex hull based on 4 control points
Monday, February 18, 13
Properties of B-Spline4. Non-periodic B-spline curve C(u) passes through
the two end control points P0 and Pn.5. Each B-spline function Nk,m(t) is nonnegative
for every t, and the family of such functions sums to unity, that is ∑ Ni,k (u) = 1
6. Affine Invariance to transform a B-Spline curve, we simply
transform each control points.7. Bézier Curves Are Special Cases of B-spline
Curves
i=0
n
Monday, February 18, 13
Properties of B-Spline8. Variation Diminishing : A B-Spline curve does
not pass through any line more times than does its control polyline
Monday, February 18, 13